Hibernate oneToMany without mapping or joining
Is it possible to add a list of Entitys into another Entity without having a mapping or joinColumn?
What I want is something like
EntityA{
@OneToMany
List<EntityX> allEntitiesX
}
This List should behave like
Query query=session.createQuery("from EntityX");
List allEntitiesX=query.list();
So if I load the EntiyA I also have acces to all other entities EntityX(with lazy-loading) without "leaving" EntityA. Due to our legacy programm I can not add a new table to connect both Entitys, because I would have to update all existing customer data and teach the old system to handle this joinTable the same way.
If possible beeing able to filter with conditions would be good aswell.
Why I want it:
In our system every "mainEntity" can have properties (EntityX).
Every property entity has a list of values, and these values are mapped with a joinTable to the "mainEntity".
If I want to add a method setProperty(propertyName, value) I have the problem that I cant find the correct propertyEntity to add my value.
The setter would have to be moved to a place where I can do entityManager querys.
hibernate jpa
add a comment |
Is it possible to add a list of Entitys into another Entity without having a mapping or joinColumn?
What I want is something like
EntityA{
@OneToMany
List<EntityX> allEntitiesX
}
This List should behave like
Query query=session.createQuery("from EntityX");
List allEntitiesX=query.list();
So if I load the EntiyA I also have acces to all other entities EntityX(with lazy-loading) without "leaving" EntityA. Due to our legacy programm I can not add a new table to connect both Entitys, because I would have to update all existing customer data and teach the old system to handle this joinTable the same way.
If possible beeing able to filter with conditions would be good aswell.
Why I want it:
In our system every "mainEntity" can have properties (EntityX).
Every property entity has a list of values, and these values are mapped with a joinTable to the "mainEntity".
If I want to add a method setProperty(propertyName, value) I have the problem that I cant find the correct propertyEntity to add my value.
The setter would have to be moved to a place where I can do entityManager querys.
hibernate jpa
It is completely unclear what you are asking.
– Alan Hay
Nov 20 '18 at 9:38
add a comment |
Is it possible to add a list of Entitys into another Entity without having a mapping or joinColumn?
What I want is something like
EntityA{
@OneToMany
List<EntityX> allEntitiesX
}
This List should behave like
Query query=session.createQuery("from EntityX");
List allEntitiesX=query.list();
So if I load the EntiyA I also have acces to all other entities EntityX(with lazy-loading) without "leaving" EntityA. Due to our legacy programm I can not add a new table to connect both Entitys, because I would have to update all existing customer data and teach the old system to handle this joinTable the same way.
If possible beeing able to filter with conditions would be good aswell.
Why I want it:
In our system every "mainEntity" can have properties (EntityX).
Every property entity has a list of values, and these values are mapped with a joinTable to the "mainEntity".
If I want to add a method setProperty(propertyName, value) I have the problem that I cant find the correct propertyEntity to add my value.
The setter would have to be moved to a place where I can do entityManager querys.
hibernate jpa
Is it possible to add a list of Entitys into another Entity without having a mapping or joinColumn?
What I want is something like
EntityA{
@OneToMany
List<EntityX> allEntitiesX
}
This List should behave like
Query query=session.createQuery("from EntityX");
List allEntitiesX=query.list();
So if I load the EntiyA I also have acces to all other entities EntityX(with lazy-loading) without "leaving" EntityA. Due to our legacy programm I can not add a new table to connect both Entitys, because I would have to update all existing customer data and teach the old system to handle this joinTable the same way.
If possible beeing able to filter with conditions would be good aswell.
Why I want it:
In our system every "mainEntity" can have properties (EntityX).
Every property entity has a list of values, and these values are mapped with a joinTable to the "mainEntity".
If I want to add a method setProperty(propertyName, value) I have the problem that I cant find the correct propertyEntity to add my value.
The setter would have to be moved to a place where I can do entityManager querys.
hibernate jpa
hibernate jpa
edited Nov 20 '18 at 15:34
OME
asked Nov 20 '18 at 8:50
OMEOME
11
11
It is completely unclear what you are asking.
– Alan Hay
Nov 20 '18 at 9:38
add a comment |
It is completely unclear what you are asking.
– Alan Hay
Nov 20 '18 at 9:38
It is completely unclear what you are asking.
– Alan Hay
Nov 20 '18 at 9:38
It is completely unclear what you are asking.
– Alan Hay
Nov 20 '18 at 9:38
add a comment |
1 Answer
1
active
oldest
votes
Yes that is possible but that will create another third table but you actually dont need third table in case of onetomany and manytoone as every row of EntityX will contain one column for EntityA.
The simplest mapping you can do is like below.
EntityA{
@OneToMany(mappedBy="entityA",cascade = CascadeType.ALL,fetch = FetchType.LAZY)
List<EntityX> allEntitiesX
}
And in another entity
EntityX{
@ManyToOne
EntityA entityA;
}
Thank you for the quick answer, but I fear this will not help me because I reality it would be a manyToMany connection (but I need only one direction). Additionaly I cant change the tables because antothes non-JPA system is working on the same database. If the answer in this case is no I will mark this answer as Accepted
– OME
Nov 20 '18 at 9:27
refer this websystique.com/hibernate/…
– Alien
Nov 20 '18 at 9:30
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes that is possible but that will create another third table but you actually dont need third table in case of onetomany and manytoone as every row of EntityX will contain one column for EntityA.
The simplest mapping you can do is like below.
EntityA{
@OneToMany(mappedBy="entityA",cascade = CascadeType.ALL,fetch = FetchType.LAZY)
List<EntityX> allEntitiesX
}
And in another entity
EntityX{
@ManyToOne
EntityA entityA;
}
Thank you for the quick answer, but I fear this will not help me because I reality it would be a manyToMany connection (but I need only one direction). Additionaly I cant change the tables because antothes non-JPA system is working on the same database. If the answer in this case is no I will mark this answer as Accepted
– OME
Nov 20 '18 at 9:27
refer this websystique.com/hibernate/…
– Alien
Nov 20 '18 at 9:30
add a comment |
Yes that is possible but that will create another third table but you actually dont need third table in case of onetomany and manytoone as every row of EntityX will contain one column for EntityA.
The simplest mapping you can do is like below.
EntityA{
@OneToMany(mappedBy="entityA",cascade = CascadeType.ALL,fetch = FetchType.LAZY)
List<EntityX> allEntitiesX
}
And in another entity
EntityX{
@ManyToOne
EntityA entityA;
}
Thank you for the quick answer, but I fear this will not help me because I reality it would be a manyToMany connection (but I need only one direction). Additionaly I cant change the tables because antothes non-JPA system is working on the same database. If the answer in this case is no I will mark this answer as Accepted
– OME
Nov 20 '18 at 9:27
refer this websystique.com/hibernate/…
– Alien
Nov 20 '18 at 9:30
add a comment |
Yes that is possible but that will create another third table but you actually dont need third table in case of onetomany and manytoone as every row of EntityX will contain one column for EntityA.
The simplest mapping you can do is like below.
EntityA{
@OneToMany(mappedBy="entityA",cascade = CascadeType.ALL,fetch = FetchType.LAZY)
List<EntityX> allEntitiesX
}
And in another entity
EntityX{
@ManyToOne
EntityA entityA;
}
Yes that is possible but that will create another third table but you actually dont need third table in case of onetomany and manytoone as every row of EntityX will contain one column for EntityA.
The simplest mapping you can do is like below.
EntityA{
@OneToMany(mappedBy="entityA",cascade = CascadeType.ALL,fetch = FetchType.LAZY)
List<EntityX> allEntitiesX
}
And in another entity
EntityX{
@ManyToOne
EntityA entityA;
}
answered Nov 20 '18 at 8:57
AlienAlien
4,85331026
4,85331026
Thank you for the quick answer, but I fear this will not help me because I reality it would be a manyToMany connection (but I need only one direction). Additionaly I cant change the tables because antothes non-JPA system is working on the same database. If the answer in this case is no I will mark this answer as Accepted
– OME
Nov 20 '18 at 9:27
refer this websystique.com/hibernate/…
– Alien
Nov 20 '18 at 9:30
add a comment |
Thank you for the quick answer, but I fear this will not help me because I reality it would be a manyToMany connection (but I need only one direction). Additionaly I cant change the tables because antothes non-JPA system is working on the same database. If the answer in this case is no I will mark this answer as Accepted
– OME
Nov 20 '18 at 9:27
refer this websystique.com/hibernate/…
– Alien
Nov 20 '18 at 9:30
Thank you for the quick answer, but I fear this will not help me because I reality it would be a manyToMany connection (but I need only one direction). Additionaly I cant change the tables because antothes non-JPA system is working on the same database. If the answer in this case is no I will mark this answer as Accepted
– OME
Nov 20 '18 at 9:27
Thank you for the quick answer, but I fear this will not help me because I reality it would be a manyToMany connection (but I need only one direction). Additionaly I cant change the tables because antothes non-JPA system is working on the same database. If the answer in this case is no I will mark this answer as Accepted
– OME
Nov 20 '18 at 9:27
refer this websystique.com/hibernate/…
– Alien
Nov 20 '18 at 9:30
refer this websystique.com/hibernate/…
– Alien
Nov 20 '18 at 9:30
add a comment |
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It is completely unclear what you are asking.
– Alan Hay
Nov 20 '18 at 9:38